Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.6 National olympiad Prove it Japan

Let $ABCD$ be a cyclic quadrilateral with an incircle of radius $6$ . Let the extensions of sides $AB$ and $DC$ beyond $B$ and $C$ , respectively, meet at $P$ , and let the extensions of sides $AD$ and $BC$ beyond $D$ and $C$ , respectively, meet at $Q$ . The inradii of triangle $PBC$ and $QCD$ are $5$ and $3$ , respectively. Find $\frac{BC}{CD}$ .

Solution

$\boxed{\frac{15}{11}}$

Let the inscribed circle of quadrilateral $ABCD$ be tangent to sides $AB$ , $BC$ , $CD$ , and $DA$ at points $S$ , $T$ , $U$ , and $V$ , respectively.
Since $AS$ and $AV$ are tangent segments from $A$ to this incircle, we have $AS = AV$ . Similarly, $BS = BT$ , $CT = CU$ and $DU = DV$ hold. Thus we have
$AB + CD = AS + BS + CU + DU = AV + BT + CT + DV = AD + BC. \quad (*)$
Since quadrilateral $ABCD$ is concyclic, we have $\angle PCB = \angle PAD$ . Therefore, triangles $PBC$ and $PDA$ are similar. Since the similarity ratio is same as the ratio of the radii of their inscribed circles, it follows that $BC : DA = 5 : 6$ . Similarly we have $CD : AB = 3 : 6 = 1 : 2$ .
Now we can write $BC = 5y$ , $DA = 6y$ , $CD = x$ and $AB = 2x$ . From $(*)$ we have $2x + x = 6y + 5y$ and thus $x : y = 11 : 3$ . We conclude that $\frac{BC}{CD} = \frac{5y}{x} = \frac{5 \cdot 3}{11} = \frac{15}{11}$ .

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